c - 学习链表

Question

对于我正在进行的这个实验，当我打印列表时，顺序和年龄是正确的，但每个元素的名称都是“quit”。知道如何解决这个问题吗？我知道它很长，但其中一些可能对我的问题至关重要(不确定是哪个)。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct person {
    char *name;
    int age;
    struct person *next;
} Person;

Person *addFront(Person *List, char *Name, int Age) {// add at front, for 
people under 21

    Person *ptrNew=malloc(sizeof(Person));
    ptrNew->name=Name;
    ptrNew->age=Age;
    ptrNew->next=List;
    return ptrNew; 
}
Person *addRear(Person *List, char *Name, int Age) { // add at rear, for 
people 21 or older

    Person *ptrNew=malloc(sizeof(Person));
    ptrNew->name=Name;
    ptrNew->age=Age;
    ptrNew->next=List;
    if (List==NULL) return ptrNew;
    Person *ptr;
    for (ptr=List; ptr != NULL; ptr = ptr->next) {
            if (ptr->next==NULL) break;
    }
    ptr->next=ptrNew;
    return List; 
}
void print(Person *List) {              // print the list (name and age for 
each item)
    Person *ptr;
    for (ptr=List; ptr != NULL; ptr = ptr->next) {
            printf("Name:%s Age:%d\n", ptr->name, ptr->age);
    }
}
void printLast(Person *List) {              // print the last person (and 
age) in the list
    Person *ptr;
        for (ptr=List; ptr != NULL; ptr = ptr->next)
            if (ptr->next==NULL) printf("Name:%s Age:%d\n", ptr->name, ptr-
>age);
}
void printFirst(Person *List) {              // print the first person (and 
age) in the list
    Person *ptr=List;
    printf("Name:%s Age:%d\n", ptr->name, ptr->age);
}
int size(Person *List) {                     // return the length of the 
list
    Person *ptr;
    int count=0;
    for (ptr=List; ptr != NULL; ptr = ptr->next) count++;
    return count;
}
int inList(Person *List, char *Name) {           // returns 1 if the name is 
in the list, else 0
    Person *ptr;
    for (ptr=List; ptr != NULL; ptr = ptr->next) {
            if (strcmp(Name, ptr->name)==0) return 1;
            else return 0;
    }

}
int getAge(Person *List, char *Name) {           // returns the age of the 
person specified
    Person *ptr;
    for (ptr=List; ptr != NULL; ptr = ptr->next) {
            if (strcmp(Name, ptr->name)==0) printf("%d", ptr->age);
            else return -1;
    }

}
                          // return -1 if the person is not in the list
int main(void) {
Person *myList = NULL;
int theAge;
char theName[128];
printf("Enter the name of a person and an age (an integer) : ");
scanf("%s %d", theName, &theAge);
while ( strcmp(theName, "quit") != 0) {
    if (theAge < 21)
        myList = addFront(myList, theName, theAge);
    else
        myList = addRear(myList, theName, theAge);
    printf("Enter another name and age (or \"quit\" and any integer when done) : ");
    scanf("%s %d", theName, &theAge);
}
printf("\n\n\nThe list is " );  print(myList);
printf("\n\nThe list has %d elements\n\n", size(myList) );
printf("\nThe first person in the list (and their age) is : "); printFirst(myList);
printf("\nThe last person in the list (and their age) is :  "); printLast(myList);
printf("\n\n");
printf("Enter the name of a person (or \"exit\" to exit) : ");
scanf("%s", theName);
while ( strcmp(theName, "exit") != 0) {
    if ( inList(myList, theName) )
        printf("\tFound %s (age is %d)\n", theName, getAge(myList, theName) );
    else
        printf("\t%s was not found in the list\n", theName);
    printf("\nEnter the name of a person (or \"exit\" to exit) : ");
    scanf("%s", theName);
}
return 0;

Answer 1

在 addFront 和 addRear 你正在使用

ptrNew->name=Name;

您不能只将列表中 name 的指针分配给您从 scanf 读取的指针。在这种情况下发生的情况是每个 List->name 都将指向相同的位置，并且如果 Name (来自 scanf)发生变化(在您的情况下为“退出”最后一个元素。每个 List->name 都会给出“quit”

您需要做的是为每个名称分别分配内存。

ptrNew->name=malloc(strlen(Name)+1);
strcpy(ptrNew->name, Name);

或者，您可以更改结构定义，以足够长度的名称存储数组

typedef struct person {
    char name[20];
    int age;
    struct person *next;
} Person;    

在这种情况下，您可以跳过上面的 malloc 步骤，但仍然需要 strcpy。